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Answer :
To calculate the work done by the system in this expansion process, we can use the equation:
Work = ∫PdV
where P is the pressure and dV is the change in volume.
From the given data, we can see that the initial pressure, P1, is 100 kPa and the final pressure, P2, is 14.4 kPa. We also have the corresponding initial volume, V1, and final volume, V2, but we don't need those values directly.
To calculate the work, we need to integrate the pressure-volume relationship over the given range. In this case, the relationship is linear, as we can see that the pressure decreases with increasing volume.
We can divide the pressure range into two segments: from 0.1 to 0.2 (ΔV1) and from 0.2 to 0.4 (ΔV2).
For the first segment, we can calculate the work done as:
Work1
= ∫PdV
= ∫(100 + m(V - 0.1))dV
where m is the slope of the line, which can be calculated as:
m = (P2 - P1) / (V2 - V1)
Substituting the values, we get:
m = (14.4 - 100) / (0.4 - 0.1
= -285.6 kPa/m^3
Now, we can integrate the equation for the first segment:
Work1
= ∫(100 - 285.6(V - 0.1))dV
= 100V - 285.6(V^2/2 - 0.1V) evaluated from 0.1 to 0.2
Evaluating the integral, we get:
Work1 = [100V - 285.6(V^2/2 - 0.1V)] from 0.1 to 0.2
Work1 = [100(0.2) - 285.6(0.2^2/2 - 0.1(0.2))] - [100(0.1) - 285.6(0.1^2/2 - 0.1(0.1))]
Work1 = 12.8 - 10.44 = 2.36 kJ
For the second segment, we can use the same approach. The only difference is the range of integration:
Work2 = ∫PdV
= ∫(14.4)dV
= 14.4∫dV
= 14.4(V2 - V1)
Substituting the values, we get:
Work2 = 14.4(0.4 - 0.2)
= 2.88 kJ
Finally, we can calculate the total work done by adding the work done in the two segments:
Total Work
= Work1 + Work2
= 2.36 kJ + 2.88 kJ
= 5.24 kJ
Therefore, the work done by the system in this expansion process is 5.24 kJ.
To learn more about expansion visit here :
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